A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be dist

Q: Why do we use $(2n+1)f_0$ for overtones in a closed pipe?

A closed pipe only supports odd harmonics because there must be a node at the closed end and an antinode at the open end. The $n$-th overtone corresponds to the $(2n+1)$-th harmonic.

Q: What is the difference between harmonics and overtones?

Harmonics are all integer multiples of the fundamental frequency. Overtones are the actual higher frequencies produced by the system. In a closed pipe, the 1st overtone is the 3rd harmonic.

Q: Why did we take the integer part of 6.16?

The number of an overtone, $n$, represents a specific physical mode of vibration, so it must be a whole number. We can't have a fraction of an overtone.

Q: What if the pipe was open at both ends?

An open pipe supports all harmonics, so the $n$-th overtone would have a frequency of $(n+1)f_0$.

Q: Is the fundamental frequency considered an overtone?

No, the fundamental frequency is the lowest resonant frequency (0th overtone). Overtones are the frequencies above the fundamental.

Question

A closed organ pipe has a fundamental frequency of $1.5 	ext{ kHz}$. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is $20,000 	ext{ Hz}$)

Options:

(A) 7

(B) 4

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

The problem of finding the number of audible overtones in a closed organ pipe is a beautiful intersection of wave physics and human biology. Let's dive into the mechanics of this acoustic system.

### Analyzing the Setup

Imagine a closed organ pipe—a tube that is open at one end and closed at the other. When air is blown into it, standing waves are formed. The simplest wave pattern, known as the fundamental mode, has a node at the closed end and an antinode at the open end.

In our problem, this fundamental frequency is given as $f_0 = 1.5 	ext{ kHz}$, which is equivalent to $1500 	ext{ Hz}$. This is the lowest pitch the pipe can produce.

### The Master Equation

But the pipe doesn't just sing one note. It produces a rich spectrum of higher frequencies called overtones. Because of the boundary conditions (a node at one end and an antinode at the other), a closed pipe only supports odd harmonics.

The frequency of the $n$-th overtone is given by the formula:
$$f_n = (2n + 1)f_0$$

Here, $n = 1$ represents the first overtone (which is the 3rd harmonic), $n = 2$ is the second overtone (5th harmonic), and so on.

### The Human Limit

The question introduces a biological constraint: the human ear can only hear frequencies up to $20,000 	ext{ Hz}$. This means that for an overtone to be distinctly heard, its frequency must be less than or equal to this maximum audible limit.

We can set up a mathematical inequality to represent this physical constraint:
$$f_n \le 20000$$

Substituting our master equation into this inequality, we get:
$$(2n + 1)f_0 \le 20000$$

### Final Calculation

Now, we simply plug in the value of our fundamental frequency, $f_0 = 1500 	ext{ Hz}$:
$$(2n + 1)(1500) \le 20000$$

Let's solve for $n$. First, divide both sides by $1500$:
$$2n + 1 \le \frac{20000}{1500}$$
$$2n + 1 \le \frac{40}{3} \approx 13.33$$

Next, subtract $1$ from both sides:
$$2n \le 12.33$$

Finally, divide by $2$:
$$n \le 6.16$$

Since $n$ represents the count of overtones, it must be a whole number. The largest integer that satisfies this condition is $6$. Therefore, a person can distinctly hear exactly **6 overtones** from this closed organ pipe.

Animated Solution for Physics - Waves: A closed organ pipe has a fundamental frequency of $1.5 kHz$ . The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is $20, 000 Hz$ )

Select Answer:

Visualized Solution

$f_{0} = 1.5 kHz$

$Overtone Frequency$

$Audible Limit$

$Substitution$

$Solving for n$

$Isolating n$

$Final Answer$

$The Way Forward$

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation

The Human Limit

Final Calculation

Similar Questions

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by $100 Hz$ than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

A closed organ pipe of length $L$ and an open organ pipe contain gases of densities $ρ_{1}$ and $ρ_{2}$ respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is

An open pipe is in resonance in $2^{nd}$ harmonic with frequency $f_{1}$ . Now one end of the tube is closed and frequency is increased to $f_{2}$ such that the resonance again occurs in $n^{th}$ harmonic. Choose the correct option.

Velocity of sound in air is $320 m/s$ . A pipe closed at one end has a length of $1 m$ . Neglecting end corrections, the air column in the pipe can resonate for sound of frequency

A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency $512 Hz$ . If the tube is opened at both ends the fundamental frequency that can be excited is (in Hz)

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is $300 m/s$ , the frequency difference between the fundamental and second harmonic of this pipe is ...... Hz. [2020, 8 Jan Shift-I]

An organ pipe $P_{1}$ closed at one end vibrating in its first harmonic and another pipe $P_{2}$ open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of $P_{1}$ and $P_{2}$ is

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is dipped vertically in water, so that half of it is in water. The fundamental frequency of the air column is now

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f0​=1.5 kHz

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Overtone Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Audible Limit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward